Question
To find:
12+22+32...................+n2
Answer: Option C
:
C
Let n=1
n32=12; n33=13
Let n=2,⇒12+22=5
n32=4; n33=83
Let n=3,⇒12+22+32=14
n32=272; n33=9
Let n=4,⇒12+22+32+42=30
n32=32; n33=643
12+22+32+⋯+n2>n33
P(n):12+22+32+⋯+n2>n33
P(1) is true
Let P(k) be true.
⇒12+22+32+⋯+k2>k33
12+22+32+⋯+k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32+⋯+k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true ∀ n∈N
Was this answer helpful ?
:
C
Let n=1
n32=12; n33=13
Let n=2,⇒12+22=5
n32=4; n33=83
Let n=3,⇒12+22+32=14
n32=272; n33=9
Let n=4,⇒12+22+32+42=30
n32=32; n33=643
12+22+32+⋯+n2>n33
P(n):12+22+32+⋯+n2>n33
P(1) is true
Let P(k) be true.
⇒12+22+32+⋯+k2>k33
12+22+32+⋯+k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32+⋯+k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true ∀ n∈N
Was this answer helpful ?
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